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Mathematics · Circles

JEE Main 2025 — 22 January, Morning Shift — Question 14

A circle C of radius 2 lies in the second quadrant and touches both the coordinate axes. Let r be the radius of a circle that has centre at the point (2,5)(2,5) and intersects the circle C at exactly two points. If the set of all possible values of rr is the interval (α,β)(\alpha, \beta), then 3β−2α3 \beta-2 \alpha is equal to :

  1. Option A:

    15

    Correct
  2. Option B:

    14

  3. Option C:

    12

  4. Option D:

    10

Answer: A

Step-by-step solution

S1:(x+2)2+(y−2)2=22S_{1}:(x+2)^{2}+(y-2)^{2}=2^{2}

S2:(x−2)2+(y−5)2=r2S_{2}:(x-2)^{2}+(y-5)^{2}=r^{2}

Both circle intersect at two points

∴∣r1−r2∣<c1c2<r1+r2\therefore\left|\mathrm{r}_{1}-\mathrm{r}_{2}\right|<\mathrm{c}_{1} \mathrm{c}_{2}<\mathrm{r}_{1}+\mathrm{r}_{2}

∣r−2∣<5<2+r|\mathrm{r}-2|<5<2+\mathrm{r}

⇒3<r<7\Rightarrow 3 < r < 7 r∈(3,7)r \in (3, 7) α=3,β=7\alpha = 3, \beta = 7 3β−2α=153\beta - 2\alpha = 15
Solution figure

Answer key and solution verified before publishing.

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Exam
JEE Main 2025
Paper
22 January, Morning Shift
Subject
Mathematics
Chapter
Circles
Topic
Family of Circles
A circle C of radius 2 lies in the second quadrant and touches both… | JEE Main 2025 PYQ with Solution · DhiX AI